MHB Amplitude, Period, frequency and phase angle

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The discussion centers on determining the resulting vibration from two simultaneous vibrations: 2cos(ωt) and 3cos(ωt + π/4). The combined expression simplifies to a form involving both cosine and sine components. The transformation involves using the identities for cosine and sine to express the result in the general form n cos(ωt ± α). The final expression indicates that the values for R and α are complex and not straightforward. The thread emphasizes the need for further calculations to finalize the solution.
jenney
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HELP!

totally lost and confused with this question:
A machine is subject to two vibrations at the same time.
one vibration has the form: 2cosωt and the other vibration has the form: 3 cos(ωt+0.785). (0.785 is actually expressed as pi/4)
determine the resulting vibration and express it in the general form of: n cos(ωt±α)
 
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jenney said:
HELP!

totally lost and confused with this question:
A machine is subject to two vibrations at the same time.
one vibration has the form: 2cosωt and the other vibration has the form: 3 cos(ωt+0.785). (0.785 is actually expressed as pi/4)
determine the resulting vibration and express it in the general form of: n cos(ωt±α)

$2\cos(\omega t) + 3\cos \left(\omega t + \dfrac{\pi}{4} \right)$

$2\cos(\omega t) + 3\left[\cos(\omega t)\cos\left(\dfrac{\pi}{4}\right) - \sin(\omega t)\sin\left(\dfrac{\pi}{4}\right)\right]$

$2\cos(\omega t) + \dfrac{3\sqrt{2}}{2}\left[\cos(\omega t) - \sin(\omega t)\right]$

$\dfrac{4+3\sqrt{2}}{2}\cos(\omega t) - \dfrac{3\sqrt{2}}{2}\sin(\omega t)$note $A\cos{x} + B\sin{x} = R\cos(x - \alpha)$, where ...

$R = \sqrt{A^2+B^2}$ and $\alpha = \arctan\left(\dfrac{B}{A}\right)$

... see what you can do from here. Note that the values for $R$ and $\alpha$ are not "nice".
 
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